What is the first step in forming #"CFC"#'s from #"CF"_3"Br"#?
1 Answer
Jun 12, 2017
Here is what I would think the first step should be:
#"CF"_3"Br" stackrel(hnu" ")(->) "F"_3"C" cdot + cdot"Br"#
which is homolytic cleavage, since the bond was split evenly so that one electron went to each center.
The fragments then recombine in symmetric and asymmetric ways, reminiscent of radical halogenation reactions:
#2"F"_3"C"cdot -> "F"_3"C"-"CF"_3# (combination of trifluoromethyl radicals)
#2"Br"cdot -> "Br"_2# (combination of two bromine radicals)
and maybe if the homolytic cleavage was not complete, we may get...
#"F"_3"C"cdot + "Br"cdot -> "CF"_3"Br"# (reverse reaction)
